Question

What is the sum of the diagonals in a rectangle whose area is 25v3 and whose width is 5?

Answer 1

Answer:

Sum of diagonals = 20 units

Step-by-step explanation:

$A(rectangle)=25\sqrt 3\: units^2\\w(rectangle) =5\: units$

To find: Sum of diagonals

$A(rectangle)=l(rectangle)\times w(rectangle)$

$\implies l(rectangle)=\frac{A(rectangle)}{w(rectangle)}$

$\implies l(rectangle)=\frac{25\sqrt 3}{5}$

$\implies l(rectangle)={5\sqrt 3}\: units$

Let the diagonals of the rectangle be $d_1\:\&\:d_2\: units$

Diagonals of a rectangle are equal in measure.

$\implies d_1=d_2=\sqrt{l(rectangle)^2+w(rectangle)^2}$

(By Pythagoras Theorem)

$\implies d_1=d_2=\sqrt{(5\sqrt 3)^2+(5)^2}$

$\implies d_1=d_2=\sqrt{75+25}$

$\implies d_1=d_2=\sqrt{100}$

$\implies d_1=d_2=10\: units$

$Sum \:of \:diagonals = d_1+d_2$

$\implies Sum \:of \:diagonals = 10+10$

$\implies Sum \:of \:diagonals = 20\: units$